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For three distinct postive integer $x,y,z$ such

$$(x+y)(x+z)=(y+z)^2$$

show that

$$(y-z)^2>8(y+z)$$

My some idea: since $$x^2+(y+z)x+yz-(y+z)^2=0$$ so $$\Delta_{x}=(y+z)^2-4yz+4(y+z)^2=5(y+z)^2-4yz=m^2$$then I can't

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  • they have to be different integers, right? Since when $ x = y = z = 1 $ then $ 2 \cdot 2 = (x+y)(x+z) = (x+z)^2 = 2^2 $ but $ (y-z)^2 = 0^2 \not > 16 = 8(y+z) $ – tmrlvi Jan 03 '15 at 12:39
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    @tmrlvi, this is said in the question. – Alex Silva Jan 03 '15 at 12:41
  • possible duplicate of http://math.stackexchange.com/questions/685432/if-distinct-numbers-a-b-c-in-mathbb-n-satisfy-abac-bc2-prove-th – r9m Jan 04 '15 at 10:57

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WLOG $y<z$, so let $z=y+a$, where $a>0$. Then the discriminant is \begin{align*}5(y+z)^2-4yz&=5(2y+a)^2-4y(y+a)=5(4y^2+4ya+a^2)-4y^2-4ya\\ &=4(4y^2+4ya+a^2)+a^2=4(2y+a)^2+a^2=(4y+2a)^2+a^2\end{align*} So if it's a square, then it's either the next square $$(4y+2a+1)^2=(4y+2a)^2+2(4y+2a)+1$$ and $$a^2=4(2y+a)+1\implies 8y+5=a^2-4a+4=(a-2)^2,$$ but that's not possible since squares cannot be $5$ mod $8$ (the remainders can only be $0$, $1$, or $4$).

Or it must be at least $$(4y+2a+2)^2=(4y+2a)^2+4(4y+2a)+4,$$ then $$(z-y)^2=a^2>4(4y+2a)=8(2y+a)=8(y+z).$$

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